I'm helping a veterinary clinic measuring pressure under a dogs paw. I use Python for my data analysis and now I'm stuck trying to divide the paws into (anatomical) subregions.

I made a 2D array of each paw, that consists of the maximal values for each sensor that has been loaded by the paw over time. Here's an example of one paw, where I used Excel to draw the areas I want to 'detect'. These are 2 by 2 boxes around the sensor with local maxima's, that together have the largest sum.

So I tried some experimenting and decide to simply look for the maximums of each column and row (can't look in one direction due to the shape of the paw). This seems to 'detect' the location of the separate toes fairly well, but it also marks neighboring sensors.

So what would be the best way to tell Python which of these maximums are the ones I want?

Note: The 2x2 squares can't overlap, since they have to be separate toes!

Also I took 2x2 as a convenience, any more advanced solution is welcome, but I'm simply a human movement scientist, so I'm neither a real programmer or a mathematician, so please keep it 'simple'.

Results

So I tried @jextee's solution (see the results below). As you can see, it works very on the front paws, but it works less well for the hind legs.

More specifically, it can't recognize the small peak that's the fourth toe. This is obviously inherent to the fact that the loop looks top down towards the lowest value, without taking into account where this is.

Would anyone know how to tweak @jextee's algorithm, so that it might be able to find the 4th toe too?

Since I haven't processed any other trials yet, I can't supply any other samples. But the data I gave before were the averages of each paw. This file is an array with the maximal data of 9 paws in the order they made contact with the plate.

This image shows how they were spatially spread out over the plate.

Update:

New update:

So after the help I got with my questions regarding paw detection and paw sorting, I was finally able to check the toe detection for every paw! Turns out, it doesn't work so well in anything but paws sized like the one in my own example. Off course in hindsight, it's my own fault for choosing the 2x2 so arbitrarily.

Here's a nice example of where it goes wrong: a nail is being recognized as a toe and the 'heel' is so wide, it gets recognized twice!

The paw is too large, so taking a 2x2 size with no overlap, causes some toes to be detected twice. The other way around, in small dogs it often fails to find a 5th toe, which I suspect is being caused by the 2x2 area being too large.

So clearly I need to change it. My own guess was changing the size of the neighborhood to something smaller for small dogs and larger for large dogs. But generate_binary_structure wouldn't let me change the size of the array.

Therefore, I'm hoping that anyone else has a better suggestion for locating the toes, perhaps having the toe area scale with the paw size?

import numpy as np
from scipy.ndimage.filters import maximum_filter
from scipy.ndimage.morphology import generate_binary_structure, binary_erosion
import matplotlib.pyplot as pp
#for some reason I had to reshape. Numpy ignored the shape header.
paws_data = np.loadtxt("paws.txt").reshape(4,11,14)
#getting a list of images
paws = [p.squeeze() for p in np.vsplit(paws_data,4)]
def detect_peaks(image):
"""
Takes an image and detect the peaks usingthe local maximum filter.
Returns a boolean mask of the peaks (i.e. 1 when
the pixel's value is the neighborhood maximum, 0 otherwise)
"""
# define an 8-connected neighborhood
neighborhood = generate_binary_structure(2,2)
#apply the local maximum filter; all pixel of maximal value
#in their neighborhood are set to 1
local_max = maximum_filter(image, footprint=neighborhood)==image
#local_max is a mask that contains the peaks we are
#looking for, but also the background.
#In order to isolate the peaks we must remove the background from the mask.
#we create the mask of the background
background = (image==0)
#a little technicality: we must erode the background in order to
#successfully subtract it form local_max, otherwise a line will
#appear along the background border (artifact of the local maximum filter)
eroded_background = binary_erosion(background, structure=neighborhood, border_value=1)
#we obtain the final mask, containing only peaks,
#by removing the background from the local_max mask
detected_peaks = local_max - eroded_background
return detected_peaks
#applying the detection and plotting results
for i, paw in enumerate(paws):
detected_peaks = detect_peaks(paw)
pp.subplot(4,2,(2*i+1))
pp.imshow(paw)
pp.subplot(4,2,(2*i+2) )
pp.imshow(detected_peaks)
pp.show()

All you need to do after is use scipy.ndimage.measurements.label on the mask to label all distinct objects. Then you'll be able to play with them individually.

Note that the method works well because the background is not noisy. If it were, you would detect a bunch of other unwanted peaks in the background. Another important factor is the size of the neighborhood. You will need to adjust it if the peak size changes (the should remain roughly proportional).

Output without overlapping squares. It seems that the same areas are selected as in your example.

Some comments

The tricky part is to calculate sums of all 2x2 squares. I assumed you need all of them, so there might be some overlapping. I used slices to cut the first/last columns and rows from the original 2D array, and then overlapping them all together and calculating sums.

When you have the sums over 2x2 squares, you can use max to find the maximum, or sort, or sorted to find the peaks.

To remember positions of the peaks I couple every value (the sum) with its ordinal position in a flattened array (see zip). Then I calculate row/column position again when I print the results.

Notes

I allowed for the 2x2 squares to overlap. Edited version filters out some of them such that only non-overlapping squares appear in the results.

Choosing fingers (an idea)

Another problem is how to choose what is likely to be fingers out of all the peaks. I have an idea which may or may not work. I don't have time to implement it right now, so just pseudo-code.

I noticed that if the front fingers stay on almost a perfect circle, the rear finger should be inside of that circle. Also, the front fingers are more or less equally spaced. We may try to use these heuristic properties to detect the fingers.

Pseudo code:

select the top N finger candidates (not too many, 10 or 12)
consider all possible combinations of 5 out of N (use itertools.combinations)
for each combination of 5 fingers:
for each finger out of 5:
fit the best circle to the remaining 4
=> position of the center, radius
check if the selected finger is inside of the circle
check if the remaining four are evenly spread
(for example, consider angles from the center of the circle)
assign some cost (penalty) to this selection of 4 peaks + a rear finger
(consider, probably weighted:
circle fitting error,
if the rear finger is inside,
variance in the spreading of the front fingers,
total intensity of 5 peaks)
choose a combination of 4 peaks + a rear peak with the lowest penalty

This is a brute-force approach. If N is relatively small, then I think it is doable. For N=12, there are C_12^5 = 792 combinations, times 5 ways to select a rear finger, so 3960 cases to evaluate for every paw.

He'll have to filter out the paws manually, given your result list ... picking the four topmost results will give him the four possibilities to construct a 2x2 square containing the maximum value 6.8
–
Johannes CharraSep 10 '10 at 14:22

The 2x2 boxes can't overlap, since if I want to do statistics, I don't want to be using the same region, I want to compare regions :-)
–
Ivo FlipseSep 10 '10 at 14:26

1

I tried it out and it seems to work for the front paws, but less so for the hind ones. Guess we'll have to try something that knows where to look
–
Ivo FlipseSep 10 '10 at 17:22

1

I explained my idea how fingers can be detected in pseudo code. If you like it, I may try to implement it tomorrow evening.
–
sastaninSep 10 '10 at 23:23

Here's a rough and ready approach, "the dumbest thing that could possibly work":

Start with five toe coordinates in roughly the place you expect.

With each one, iteratively climb to the top of the hill. i.e. given current position, move to maximum neighbouring pixel, if its value is greater than current pixel. Stop when your toe coordinates have stopped moving.

To counteract the orientation problem, you could have 8 or so initial settings for the basic directions (North, North East, etc). Run each one individually and throw away any results where two or more toes end up at the same pixel. I'll think about this some more, but this kind of thing is still being researched in image processing - there are no right answers!

Assign each pixel to the closest cluster (just make a list for each cluster).

Calculate the center of mass of each cluster. For each cluster, this is: Sum(coordinate * intensity value)/Sum(coordinate)

Move each cluster to the new centre of mass.

This method will almost certainly give much better results, and you get the mass of each cluster which may help in identifying the toes.

(Again, you've specified the number of clusters up front. With clustering you have to specify the density one way or another: Either choose the number of clusters, appropriate in this case, or choose a cluster radius and see how many you end up with. An example of the latter is mean-shift.)

Sorry about the lack of implementation details or other specifics. I would code this up but I've got a deadline. If nothing else has worked by next week let me know and I'll give it a shot.

The problem is, that the paws change their orientation and I don't have any calibration/baseline of a correct paw to start with. Plus I fear that a lot of the image recognition algorithms are a bit out of my league.
–
Ivo FlipseSep 10 '10 at 15:12

The "rough and ready" approach is pretty simple - maybe I didn't the idea well. I'll put in some pseudocode to illustrate.
–
CakeMasterSep 10 '10 at 16:14

I have a feeling your suggestion will help fix the recognition of the hind paws, I just don't know 'how'
–
Ivo FlipseSep 10 '10 at 17:31

I've added another idea. By the way if you have a load of good data it would be cool to put it online somewhere. It could be useful for people studying image processing / machine learning and you might get some more code out of it...
–
CakeMasterSep 10 '10 at 19:19

1

I was just thinking about writing down my data processing on a simple Wordpress blog, simply to be of use for other and I have to write it down anyway. I like all your suggestions, but I fear I'll have to wait for someone without a deadline ;-)
–
Ivo FlipseSep 10 '10 at 19:22

Cool, you make my work sound like rocket science!
–
Ivo FlipseSep 10 '10 at 23:56

For glancing over the article it does seem to describe the same data processing as what I'm trying here, however I fear it greatly surpassed my programming skills :(
–
Ivo FlipseSep 10 '10 at 23:59

@Ivo: I've never tried to implement it myself. I just use ROOT. There are python bindings, none-the-less, but be aware that ROOT is a pretty heavy package.
–
dmckeeSep 11 '10 at 0:13

@Ivo Flipse: I agree with dmckee. You have a lot of promising leads in other answers. If they all fail and you feel like investing some time, you can delve into ROOT and it will (probably) do what you need it to. I've never known anyone who's tried to learn ROOT via the python bindings (rather than it's natural C++), so I wish you luck.
–
physicsmichaelSep 11 '10 at 22:55

With gradient, you mean I should calculate the steepness of the slopes, once this get's above a certain value I know there's 'a peak'? I tried this, but some of the toes only have very low peaks (1.2 N/cm) compared to some of the others (8 N/cm). So how should I handle peaks with a very low gradient?
–
Ivo FlipseSep 10 '10 at 13:37

2

What's worked for me in the past if I couldn't use the gradient directly was to look at gradient and the maxima, e.g. if the gradient is a local extrema and I'm at a local maxima, then I'm at a point of interest.
–
ChrisCSep 10 '10 at 14:27

Here is an idea: you calculate the (discrete) Laplacian of the image. I would expect it to be (negative and) large at maxima, in a way that is more dramatic than in the original images. Thus, maxima could be easier to find.

Here is another idea: if you know the typical size of the high-pressure spots, you can first smooth your image by convoluting it with a Gaussian of the same size. This may give you simpler images to process.

Physicist's solution:
Define 5 paw-markers identified by their positions X_i and init them with random positions.
Define some energy function combining some award for location of markers in paws' positions with some punishment for overlap of markers; let's say:

E(X_i;S)=-Sum_i(S(X_i))+alfa*Sum_ij (|X_i-Xj|<=2*sqrt(2)?1:0)

(S(X_i) is the mean force in 2x2 square around X_i, alfa is a parameter to be peaked experimentally)

Now time to do some Metropolis-Hastings magic:
1. Select random marker and move it by one pixel in random direction.
2. Calculate dE, the difference of energy this move caused.
3. Get an uniform random number from 0-1 and call it r.
4. If dE<0 or exp(-beta*dE)>r, accept the move and go to 1; if not, undo the move and go to 1.
This should be repeated until the markers will converge to paws. Beta controls the scanning to optimizing tradeoff, so it should be also optimized experimentally; it can be also constantly increased with the time of simulation (simulated annealing).

Care to show how this would work on my example? As I'm really not into high level math, so I already have a hard time unraveling the formula you proposed :(
–
Ivo FlipseSep 10 '10 at 19:43

1

This is high school math, probably my notation is just obfuscated. I have a plan to check it, so stay tuned.
–
mbqSep 10 '10 at 20:08

I'm a particle physicist. For a long time the go-to software tool in our discipline was called PAW, and it had a entity related to graphs called a "marker". You can imagine how confusing I found this answer on the first couple of times around...
–
dmckeeSep 11 '10 at 0:43

thanks for the raw data. I'm on the train and this is as far as I've gotten (my stop is coming up). I massaged your txt file with regexps and have plopped it into a html page with some javascript for visualization. I'm sharing it here because some, like myself, might find it more readily hackable than python.

I think a good approach will be scale and rotation invariant, and my next step will be to investigate mixtures of gaussians. (each paw pad being the center of a gaussian).

Heres another approach that I used when doing something similar for a large telescope:

1) Search for the highest pixel.
Once you have that, search around that for the best fit for 2x2 (maybe maximizing the 2x2 sum), or do a 2d gaussian fit inside the sub region of say 4x4 centered on the highest pixel.

Then set those 2x2 pixels you have found to zero (or maybe 3x3) around the peak center

go back to 1) and repeat till the highest peak falls below a noise threshold, or you have all the toes you need

If it's worth the trouble, I wouldn't mind annotating a large sample by hand. My problem would be: how do I implement this, since I know nothing about programming neural networks
–
Ivo FlipseSep 10 '10 at 18:52

you'll have to make a decision about whether to use 4 or 8 connectedness. personally, for most problems i prefer 6-connectedness. anyway, once you've separated out each "paw print" as a connected region, it should be easy enough to iterate through the region and find the maxima. once you've found the maxima, you could iteratively enlarge the region until you reach a predetermined threshold in order to identify it as a given "toe".

one subtle problem here is that as soon as you start using computer vision techniques to identify something as a right/left/front/rear paw and you start looking at individual toes, you have to start taking rotations, skews, and translations into account. this is accomplished through the analysis of so-called "moments". there are a few different moments to consider in vision applications:

Interesting, also your point about rotation is important, as the sensors of the pressure plate are rectangular instead of square. Which has a significant impact if the paw is significantly rotated
–
Ivo FlipseSep 10 '10 at 19:33

I basically just make an array with the position of the upper-left and the sum of each 2x2 square and sort it by the sum. I then take the 2x2 square with the highest sum out of contention, put it in the best array, and remove all other 2x2 squares that used any part of this just removed 2x2 square.

It seems to work fine except with the last paw (the one with the smallest sum on the far right in your first picture), it turns out that there are two other eligible 2x2 squares with a larger sum (and they have an equal sum to each other). One of them is still selects one square from your 2x2 square, but the other is off to the left. Fortunately, by luck we see to be choosing more of the one that you would want, but this may require some other ideas to be used to get what you actually want all of the time.

Apply a low pass filter, such as convolution with a 2D gaussian mask. This will give you a bunch of (probably, but not necessarily floating point) values.

Perform a 2D non-maximal suppression using the known approximate radius of each paw pad (or toe).

This should give you the maximal positions without having multiple candidates which are close together. Just to clarify, the radius of the mask in step 1 should also be similar to the radius used in step 2. This radius could be selectable, or the vet could explicitly measure it beforehand (it will vary with age/breed/etc).

Some of the solutions suggested (mean shift, neural nets, and so on) probably will work to some degree, but are overly complicated and probably not ideal.

I'm sure you have enough to go on by now, but I can't help but suggest using the k-means clustering method. k-means is an unsupervised clustering algorithm which will take you data (in any number of dimensions - I happen to do this in 3D) and arrange it into k clusters with distinct boundaries. It's nice here because you know exactly how many toes these canines (should) have.

I actually made a list with all the 2x2 sums, but when I had them ordered I had no idea how to iteratively compare them. My problem was that when I sorted it, I lost track of the coordinates. Perhaps I could stick them in a dictionary, with the coordinates as the key.
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Ivo FlipseSep 10 '10 at 13:08

Yes, some kind of dictionary would be necessary. I would have assumed that your representation of the grid is some sort of dictionary already.
–
Johannes CharraSep 10 '10 at 13:31

Well the image you see above is a numpy array. The rest is currently stored in multidimensional lists. It would probably be better to stop doing that, though I'm not as familiar with iterating over dictionaries
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Ivo FlipseSep 10 '10 at 13:53

Hmmm that setting to zero would at least remove it from any further calculations, that would be useful.
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Ivo FlipseSep 10 '10 at 13:10

Instead of setting to zero, you may calculate a gaussian function with hand picked parameters and subtract the found values from the original pressure readings. So if the toe is pressing your sensors, then by finding the highest pressing point, you use it to decrease the effect of that toe on the sensors, thus, eliminating the neighbouring cells with high pressure values. en.wikipedia.org/wiki/File:Gaussian_2d.png
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DaniyarSep 10 '10 at 17:48

Care to show an example based on my sample data @Daniyar? As I'm really not familiar with such kind of data processing
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Ivo FlipseSep 10 '10 at 18:39